powering clock from 12 volt battery?
I have a little clock that's powered with a AA battery , How would I go about wiring it so that it is powered from a 12 volt car battery? Thanks
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I had to do this for an aircraft - the clock was designed to operate on +5 Vdc and the aircraft battery was +24 Vdc. The way I solved it was simply to add an inline series resistor; total cost of 29 cents.
The clock consumes a very small but fairly consistent amount of power, probably something around 1 to 2 milliamps (that's one-thousandth of an amp) - to make the math easy, let's assume 1.5 mA. With a battery voltage of roughly +1.5 Vdc, we're looking at an equivalent resistance for the clock at 1,000 ohms using Ohm's law (i.e., R=V/I, or Resistance equals Voltage divided by Current).
So.... if you want to drop the +12 Vdc of your vehicle electrical system down to +1.5 Vdc, you need to install a series resistor in between the battery voltage and the clock of roughly 10 times the clock resistance, or 10,000 ohms (10 Kohms).
If you don't have the ability to accurately measure the actual current draw, then you won't be able to calculate the exact value of resistor needed, but 10 Kohms will probably get you into the ballpark.
A cheap DVM would allow you to nail it precisely, even if it won't accurately measure current way down in the milliamp range. Simply put the 10 Kohm resistor in series with the clock (i.e., connect the resistor between the +12V terminal and the red or plus lead of the clock, then connect the black or negative lead of clock to GND, or ground).
Next measure the voltage between the positive and negative leads for the clock - if it's less than +1.5 Vdc, then you will need to DECREASE the value of the dropping resistor, if it's more than +1.5 Vdc, then you will need to INCREASE the value of the dropping resistor.
Please note that the clock is probably fairly tolerant towards the voltage input, so anything between 1 to 2 volts would likely work OK.
Dale
The clock consumes a very small but fairly consistent amount of power, probably something around 1 to 2 milliamps (that's one-thousandth of an amp) - to make the math easy, let's assume 1.5 mA. With a battery voltage of roughly +1.5 Vdc, we're looking at an equivalent resistance for the clock at 1,000 ohms using Ohm's law (i.e., R=V/I, or Resistance equals Voltage divided by Current).
So.... if you want to drop the +12 Vdc of your vehicle electrical system down to +1.5 Vdc, you need to install a series resistor in between the battery voltage and the clock of roughly 10 times the clock resistance, or 10,000 ohms (10 Kohms).
If you don't have the ability to accurately measure the actual current draw, then you won't be able to calculate the exact value of resistor needed, but 10 Kohms will probably get you into the ballpark.
A cheap DVM would allow you to nail it precisely, even if it won't accurately measure current way down in the milliamp range. Simply put the 10 Kohm resistor in series with the clock (i.e., connect the resistor between the +12V terminal and the red or plus lead of the clock, then connect the black or negative lead of clock to GND, or ground).
Next measure the voltage between the positive and negative leads for the clock - if it's less than +1.5 Vdc, then you will need to DECREASE the value of the dropping resistor, if it's more than +1.5 Vdc, then you will need to INCREASE the value of the dropping resistor.
Please note that the clock is probably fairly tolerant towards the voltage input, so anything between 1 to 2 volts would likely work OK.
Dale
